and {$p,q$} are chosen to satisfy the Pell equation $p^2-85q^2 =\mp 1$. (Actually, $p,q$ are half-integers since one can use $p^2-85q^2 =\mp 4$.) Ramanujan missed using {$p,-q$} which yields the different,

and {$v_1,\, v_2,\, w$} = {$c^2-d^2,\; a^2-b^2,\; (a+b)(c+d)$}. For example, using {$a,b,c,d$} = {$6,-9,8,1$}, and tweaking (5) with a simple linear transformation so the last term transforms from $x^2-45xy+216y^2$ to $p^2-321q^2$ yields,

thus a sum of cubes identity analogous to Ramanujan's. (Using {$p,-q$} will yield the second family.)

Question: Starting with an initial {$a,b,c,d$}, especially the form {$a,b,c,\pm1$}, and using the quadratic parametrization (5) where one term has been equated to $\pm1$, is it always possible to find a generating function analogous to (1)? (I've tried various {$a,b,c,d$} and it seems to be always the case, but a general proof eludes me.) See also Rowlands' survey article here.

1 Answer
1

Yes, this is true. As in your example, you need to start with integers $x,y$ such that the last term of the LHS of (5) is $dx^2+v_2xy+cwy^2=\pm1$. Make a change of variables to eliminate the $xy$ term (i.e., "complete the square"), which yields an equation of the form $p^2-Dq^2=1$. When $D$ is a positive nonsquare, this says that the element $p+q\sqrt{D}$ of $\mathbb{Z}[\sqrt{D}]$ has norm $1$. But it's classical to write down all elements of norm $1$: they are $\pm u^n$ where $n\in\mathbb{Z}$ and $u$ is the smallest element of $\mathbb{Z}[\sqrt{D}]$ such that $N(u)=1$ and $u>1$. Say
$$
p+q\sqrt{D} = u^n
$$
(the case of $-u^n$ works exactly the same way). Then $p-q\sqrt{D} = 1/u^n$, so
$$
p=\frac{u^n+u^{-n}}2 \quad\text{ and }\quad q=\frac{u^n-u^{-n}}{2\sqrt{D}}.
$$
Let's just consider nonnegative values of $n$, and write $p_n$ and $q_n$ for the above expressions. Then the corresponding values of $x$ and $y$ (call them $x_n$ and $y_n$) are linear combinations of $p_n$ and $q_n$, and hence of $u^n$ and $u^{-n}$. Writing, for instance, $a_n=ax_n^2−v_1x_ny_n+bwy_n^2$ for the value of the first number you're cubing, it follows that $a_n$ is a linear combination of $u^{2n}$, $1$, and $u^{-2n}$. Therefore
$$
\sum_{n=0}^\infty a_n x^n
$$
is a linear combination of
$$
\sum_{n=0}^\infty u^{2n} x^n,\quad \sum_{n=0}^\infty x^n, \quad\text{and}\quad\sum_{n=0}^\infty u^{-2n} x^n,
$$
or equivalently of
$$
\frac{1}{1-u^2 x},\quad \frac{1}{1-x},\quad\text{and}\quad\frac{1}{1-u^{-2}x}.
$$
Thus the generating function for $a_n$ is a rational function, and we could do the same for $b_n$ and $c_n$, in order to get a result similar to Ramanujan's.